Question: Simplify; express your answer in exponential form. Assume $x\neq 0, k\neq 0$. $\dfrac{{(x^{3}k^{-4})^{2}}}{{(x^{-3}k^{-3})^{-3}}}$
Explanation: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(x^{3}k^{-4})^{2} = (x^{3})^{2}(k^{-4})^{2}}$ On the left, we have ${x^{3}}$ to the exponent ${2}$ . Now ${3 \times 2 = 6}$ , so ${(x^{3})^{2} = x^{6}}$ Apply the ideas above to simplify the equation. $\dfrac{{(x^{3}k^{-4})^{2}}}{{(x^{-3}k^{-3})^{-3}}} = \dfrac{{x^{6}k^{-8}}}{{x^{9}k^{9}}}$ Break up the equation by variable and simplify. $\dfrac{{x^{6}k^{-8}}}{{x^{9}k^{9}}} = \dfrac{{x^{6}}}{{x^{9}}} \cdot \dfrac{{k^{-8}}}{{k^{9}}} = x^{{6} - {9}} \cdot k^{{-8} - {9}} = x^{-3}k^{-17}$